A group of n students can be divided into equal groups of 4

A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?

A. 33
B. 46
C. 49
D. 53
E. 86

Given:
\(n=4q+1\), so \(n\) could be: 1, 5, 9, 13, ...
\(n=5p+3\), so \(n\) could be: 3, 8, 13, ...

General formula for \(n\) based on above two statements will be: \(n=20m+13\) (the divisor should be the least common multiple of above two divisors 4 and 5, so 20 and the remainder should be the first common integer in above two patterns, hence 13). For more about this concept see: manhattan-remainder-problem-93752.html#p721341, when-positive-integer-n-is-divided-by-5-the-remainder-is-90442.html#p722552, when-the-positive-integer-a-is-divided-by-5-and-125591.html#p1028654

From, \(n=20m+13\) we have that the two smallest possible values of \(n\) are 13 (for \(m=0\)) and 33 (for \(m=1\)).

13+33=46.

Answer: B.

Hope it helps.
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Yes you can do the way you started by listing the possible values of n for both patterns and then picking first two matching numbers from these lists. Since we are dealing with easy and small numbers this approach probably would be the fastest one.

A group of n students can be divided into equal groups of 4 with 1 student left over --> n=4q+1 --> n can be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, ... (basically an evenly spaced set with common difference of 4)

A group of n students can be divided into equal groups of 5 with 3 students left over --> n=5p+3 --> n can be: 3, 8, 13, 18, 23, 28, 33, 38, ... (basically an evenly spaced set with common difference of 5)

Therefor two smallest possible values of n are 13 and 33 --> 13+33=46.

Answer: B.

Else you can derive general formula based on n=4q+1 and n=5p+3.

Divisor will be the least common multiple of above two divisors 4 and 5, hence 20.

Remainder will be the first common integer in above two patterns, hence 13 --> so, to satisfy both conditions, n must be of a type n=20m+13: 13, 33, 53, ... (two two smallest possible values of n are for m=0 and for m=1, so 13, and 33 respectively) --> 13+33=46.

Answer: B.

For more about this concept see:
https://gmatclub.com/forum/manhattan-re ... ml#p721341
https://gmatclub.com/forum/when-positiv ... ml#p722552
https://gmatclub.com/forum/when-the-pos ... l#p1028654

Hope it helps.
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n = 4q + 1
n = 5q + 3

I'll start with the first equation: n = 5+k4 where k = 0,1,2,3, ... etc
also, n = 8+m5 where m = 0,1,2,3,.. etc


for first equation: 5,9,13,17,21,25,29,33,37,41,45
for second equation: 8,13,18,23,28,33,38,43,48,53

The sum of minimum n's = 13 + 33 = 46


B

A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?

A. 33
B. 46
C. 49
D. 53
E. 86

OA is A and this is how I arrived at.

Let say n = 4x+1 and n = 5y+3 -----> From the question Stem

n=4x+1 n = 5y+3
5 8
9 13
13 18
17 23
25 33
29 38
33 43

I get these above values by putting the same values for x and y. Is my concept correct?

Hi Bunuel - can the values of q and p be ZERO? I don't think they can be and therefore n cannot be 1 & 3. Am I wrong?

THEORY:
Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is non-negative integer and always less than divisor).

For example we are told that when positive integer n is divided by 25, the remainder is 13 --> \(n=25q+13\). Now, the lowest value of \(q\) can be zero and in this case \(n=13\) --> 13 divided by 25 yields the remainder of 13. Generally when divisor (25 in our case) is more than dividend (13 in our case) then the reminder equals to the dividend. For example:
3 divided by 24 yields a reminder of 3 --> \(3=0*24+3\);
or:
5 divided by 6 yields a reminder of 5 --> \(5=0*6+5\).

Also note that you shouldn't worry about negative numbers in divisibility questions, as every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

OUR ORIGINAL QUESTION:
We are told that "a group of n students can be divided into equal groups of 4 with 1 student left over" --> n=4q+1. Here q also can be zero, which would mean that there is only 1 student and zero groups of 4.

QUESTIONS TO PRACTICE:
PS questions on remainders: search.php?search_id=tag&tag_id=199
DS questions on remainders: search.php?search_id=tag&tag_id=198

THEORY ON REMAINDERS: compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html

Hope it helps.
_________________

You are a true genius buddy.

n=4q+1 -----> 1,5,9,13,17,21,25,29,33

n=5q+3------->3,8,13,18,23,28,33

the first two common numbers are 13 and 33, so add those numbers, you get 13+33=46

so, asnwer is B, 46

hope this helps

Isn't there any arithmetic solution to this question. I mean, just Hit n Trial method. Indeed there must be an arithmetic way out. Using this hit and trial method sometimes takes much longer time, henceforth I needed to go with a systematic approach.

4x + 1 = n (1)
5y + 3 = n (2)

Equating (1) and (2)
4x + 1 = 5y + 3
4x = 5y + 2
Put y=1,2,3,4,etc.
Since (5y + 2) need to be a multiple of 4 to satisfy the equation on the left side. The 2 minimum values of y are 2 and 6.

So, n = 5y + 3
n = 5(2) + 3 = 13 and
n = 5(6) + 3 = 33

Adding the 2 minimum values of n
13 + 33 = 46

So, the correct answer is B.

assume difference of 1 between quotients
(n-1)/4-(n-3)/5=1
n=13=smallest n
13+4*5=33=second smallest n
13+33=46
B

Hi All,

You would likely find it easiest to 'brute force' this question (simply write down enough of the possibilities until you either spot the pattern involved or have the exact answer on your pad).

Equal groups of 4 with 1 left over COULD be... 5, 9, 13, 17, 21, 25, 29, 33.....
Equal groups of 5 with 3 left over COULD be... 8, 13, 18, 23, 28, 33....

The two SMALLEST values that fit BOTH groups are 13 and 33. We're asked for the sum of those values...

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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